Why we say that EM waves are transverse in nature? I have seen some proofs regarding my question but they all calculate flux through imaginary cube. Here is My REAL problem that I can't here imagine infinitesimal area for calculating flux because em line of force will intersect (perpendicular or not) surface at only one point so $E.ds$ will be zero so even flux through one surface of cube will always be zero. I am Bit Confused. I DON'T KNOW VECTOR CALCULUS BUT KNOW CALCULUS.
Answer
Why we say that em waves are transverse in nature?
In a region empty of electric charge, we have, from Maxwell's equations:
$$\nabla \cdot \vec E = \nabla \cdot \vec B = 0$$
Since you don't yet know vector calculus, let's rewrite these divergence equations as so:
$$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 0 $$
$$\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0 $$
Now, assume an electromagnetic wave is propagating in the $z$ direction so that the space and time variation of the field components are of the form
$$\cos(kz - \omega t)$$
Since the spatial variation is zero in the $x$ and $y$ directions, our equations become
$$\frac{\partial E_z}{\partial z} = 0$$
$$\frac{\partial B_z}{\partial z} = 0$$
Which means that electric and magnetic field components in the $z$ direction, the direction of propagation, must be constant with respect to $z$.
In other words, only the electric and magnetic field components transverse to the direction of propagation vary with respect to $z$. i.e., the electromagnetic wave is transverse.
Addendum to address a comment:
Why spatial Variations are zero in both x and y directions.
We stipulated that the field components are of the form $\cos(kz - \omega t)$ which means the wave is propagating in the $z$ direction.
Clearly, the partial derivative of $\cos(kz - \omega t)$ with respect to $x$ and $y$ is zero.
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